Uniform convergence and pointwise limit of $f_n(x)=\frac 1{1+x^n}$

Question

let $f_n(x)=\frac 1{1+x^n}$

a)Find the pointwise lim. $f(x)=lim_{n\to\infty}f_n(x),n\in[0,\infty)$

b)is $f_n$ uniformly convergent on the interval $[0,\frac 12]$?

for b) $x=0\to f_n(x)\to 1$

$x=\frac12\to f_n(x)\to 1$

$x\in (0,\frac 12)\to f_n(x)\to 1$

$f'_n( x)=\frac{-nx^{n-1}}{(1+x^{n})^2}=0\to x^{n-1}=0\to x=0$ ( since for x<0 $f'_n(x)<0,$ is $f_n(0)=1$ min??) how can I continue

for a) my guess is $f_n(x)$ pointwise converges to $f(x)=\begin{cases}1\quad, x\in[0,1)\\0\quad,x\in[1,\infty)\end{cases}$ is this correct? how can I prove?

Answer 1

Your guess for the pointwise limite is almost correct. Note that $f_n(1)=\frac{1}{1+1^n}=\frac{1}{2},$ and thus $f(1)=\lim_{n\to \infty} f_n(1)=\frac{1}{2}.$

With respect to the absolute convergence note that:

$$\left|f_n(x)-f(x)\right|=\left|\frac{1}{1+x^n}-1\right|=\frac{x^n}{1+x^n}\le x^n\le \frac{1}{2^n}, \forall x\in \left[0,\frac12\right].$$

Answer 2

The point-wise limit isn't quite correct: what's the value of $f$ at $x=1$?

For the uniform convergence notice that we can answer on this manner:

$$\sup_{x\in[0,1/2]}|f_n(x)-f(x)|=\sup_{x\in[0,1/2]}1-\frac{1}{1+x^n}=1-\frac{1}{1+\left(\frac12\right)^n}\xrightarrow{n\to\infty}0$$ and then we conclude.

Answer 3

You can check that

a) $$f(x) = \left\{\begin{matrix} 1 & 0 \leq x <1\\ \frac{1}{2}& x = 1\\ 0 & x > 1\\ \end{matrix}\right.$$

b) See user63181's answer or following to mfl's estimation, choose $N = \frac{-\ln \epsilon }{\ln 2}.$